mc-ty-circles-2009-1.pdf hshahqmavwmskwhayw

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1. Thegeometryofa
circle
mc-TY-circles-2009-1
In this unit we find the equation of a circle, when we are told its centre and its radius. There are
two different forms of the equation, and you should be able to recognise both of them. We also
look at some problems involving tangents to circles.
In order to master the techniques explained here it is vital that you undertake plenty of practice
exercises so that they become second nature.
After reading this text, and/or viewing the video tutorial on this topic, you should be able to:
•find the equation of a circle, given its centre and radius;
•find the centre and radius of a circle, given its equation in standard form;
•find the equation of the tangent to a circle through a given point on its circumference;
•decide whether a given line is tangent to a given circle.
Contents
1.Introduction 2
2.The equation of a circle centred at the origin 2
3.The general equation of a circle 4
4.The equation of a tangent to a circle at a given point 7
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2. 1. Introduction
The circle is a familiar shape and it has a host of geometric properties that can be proved using
the traditional Euclidean format. But it is sometimes useful to work in co-ordinates and this
requires us to know the standard equation of a circle, how to interpret that equation and how
to find the equation of a tangent to a circle. This video will explore these particular facets of a
circle, using co-ordinate geometry.
2. The equation of a circle centred at the origin
The simplest case is that of a circle whose centre is at the origin. Let us take an example. What
will be the equation centred on the origin with radius 5 units?
5
P(x, y)
x
y
5
5
−5
−5
NO
If we take any pointP(x, y)on the circle, thenOP= 5is the radius of the circle. ButOPis also
the hypotenuse of the right-angled triangleOPN, formed when we drop a perpendicular fromP
to thex-axis. Now in the right-angled triangle,ON=xandNP=y. Thus, using the theorem
of Pythagoras,
x
2
+y
2
= 5
2
= 25.
And this equation is true for any point on the circle. For instance, we could take a pointQ(x1, y1)
in a different quadrant.
5
Q(x
1
, y
1
)
x
1
y
1
5
5
−5
−5
N O
Once again, we can drop a perpendicular fromQto thex-axis. And now we can use the right-
angled triangleOQNto see thatx
2
1
+y
2
1
= 5
2
. So the co-ordinates(x1, y1)of the pointQalso
satisfy the equationx
2
+y
2
= 25.
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3. We shall now take the radius of the circle to ber.
r
P(x, y)
x
y
r
r
−r
−r
NO
If we take any pointP(x, y)on the circle, thenOP=ris the radius of the circle. ButOPis also
the hypotenuse of the right-angled triangleOPN, formed when we drop a perpendicular fromP
to thex-axis. In the right-angled triangle,ON=xandNP=y. Thus, using the theorem of
Pythagoras,
x
2
+y
2
=r
2
,
and this is the equation of a circle of radiusrwhose centre is the originO(0,0).
Key Point
The equation of a circle of radiusrand centre the origin is
x
2
+y
2
=r
2
.
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4. 3. The general equation of a circle
What is the equation of a circle of radiusr, centred at the pointC(a, b)?
P(x, y)
x − a
y − b
r
N
O
C(a, b)
We shall take a horizontal line through the centreCand drop a perpendicular fromPto meet
this horizontal line atN. Then again we have a right-angled triangleCPN, whereCP=ris
the hypotenuse, and where we haveCN=x−aandPN=y−b. Thus using Pythagoras again
we have
CN
2
+PN
2
=CP
2
,
so that
(x−a)
2
+ (y−b)
2
=r
2
.
Expanding the brackets gives
x
2
−2ax+a
2
+y
2
−2by+b
2
=r
2
,
and if we bringr
2
to the left-hand side and rearrange we get
x
2
−2ax+y
2
−2by+a
2
+b
2
−r
2
= 0.
It is a convention, at this point, to replace−abygand−bbyf. This gives
x
2
+ 2gx+y
2
+ 2fy+g
2
+f
2
−r
2
= 0.
Now look at the last three terms on the left-hand side,g
2
+f
2
−r
2
. These do not involvexory
at all, so together they just represent a single number that we can callc. Substituting this into
the equation finally gives us
x
2
+ 2gx+y
2
+ 2fy+c= 0.
This is the general equation of a circle. We can recognise it,because it is quadratic in bothx
andy, and it has two additional properties. First, there is no term inxy. And secondly, the
coefficient ofx
2
is the same as the coefficient ofy
2
. (In fact, our equation has both coefficients
equal to 1, but you can always multiply an equation by a non-zero constant to obtain another
valid equation, and so we must allow for this possibility.) The centre of the circle is then at
(a, b) = (−g,−f)and, sincec=g
2
+f
2
−r
2
, we have
r
2
=g
2
+f
2
−c ,
so that the radius of the circle is given by
r=
p
g
2
+f
2
−c .
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5. Key Point
The general equation of a circle is
x
2
+y
2
+ 2gx+ 2fy+c= 0,
where the centre is given by(−g,−f)and the radius byr=
p
g
2
+f
2
−c. The equation can
be recognised because it is given by a quadratic expression in bothxandywith noxyterm,
and where the coefficients ofx
2
andy
2
are equal.
Example
Find the centre and radius of the circle
x
2
+y
2
−6x+ 4y−12 = 0.
Solution
First, we can check that the expression on the left-hand sideis quadratic, that there is no term
involvingxy, and that the coefficients ofx
2
andy
2
are equal. So this is the equation of a circle.
If we compare this equation with the standard equation
x
2
+y
2
+ 2gx+ 2fy+c= 0,
we see thatg=−3andf= 2, so that the centre is(−g,−f) = (3,−2). We also see that
c=−12, so we can find the radius by calculating
r=
p
g
2
+f
2
−c
=
p
(−3)
2
+ 2
2
−(−12)
=
√
9 + 4 + 12
=
√
25
= 5.
An alternative method is to attempt to reconstruct the equation of the circle in the form
(x−a)
2
+ (y−b)
2
=r
2
by completing the square. We start by collecting together the terms inx, and the terms iny.
So we rewrite our equation
x
2
+y
2
−6x+ 4y−12 = 0
as
x
2
−6x+y
2
+ 4y−12 = 0.
Now the terms inxmust come from(x−a)
2
, a complete square, so we complete the square for
thexterms, and similarly for theyterms, to get
(x−3)
2
−9 + (y+ 2)
2
−4−12 = 0.
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6. So we have
(x−3)
2
+ (y+ 2)
2
−25 = 0,
which we rewrite as
(x−3)
2
+ (y+ 2)
2
= 25 = 5
2
.
We can now see that the centre of the circle is(3,−2)and the radius is 5.
Example
Find the centre and radius of the circle
2x
2
+ 2y
2
−8x−7y= 0.
Solution
Notice that this is the equation of a circle, even though the coefficients ofx
2
and ofy
2
are not
equal to 1. But we can divide throughout by 2, and we get
x
2
+y
2
−4x−
7
2
y= 0.
If we compare this with the standard equation
x
2
+y
2
+ 2gx+ 2fy+c= 0
we see thatg=−2andf=−
7
4
, so the centre of the circle is(−g,−f) = (2,
7
4
). We also see
thatc= 0, so we find the radius by calculating
r=
p
g
2
+f
2
−c
=
q
(−2)
2
+
−
−
7
4

2
=
q
4 +
49
16
=
q
113
16
=
1
4
√
113.
Alternatively, we could try completing the square to regainthe form(x−a)
2
+ (y−b)
2
=r
2
. So
we start with our equation
2x
2
+ 2y
2
−8x−7y= 0,
and again we divide by 2 to get
x
2
+y
2
−4x−
7
2
y= 0.
Collecting thexterms together and theyterms together, we get
x
2
−4x+y
2
−
7
2
y= 0,
and then completing the square gives us
(x−2)
2
−4 +
−
y−
7
4

2
−
49
16
= 0
so that
(x−2)
2
+
−
y−
7
4

2
=
113
16
.
We can now see that the centre of the circle is(2,
7
4
)and the radius is
1
4
√
113.
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7. Example
Find the centre and radius of the circle
x
2
+y
2
+ 8x+ 7 = 0.
Solution
Notice that, in this example, there is noyterm. If we compare our equation with the standard
equation
x
2
+y
2
+ 2gx+ 2fy+c= 0,
we see thatg= 4andf= 0. So the centre of the circle is(−g,−f) = (−4,0). We also see
thatc= 7, so we find the radius by calculating
r=
p
g
2
+f
2
−c
=
√
4
2
+ 0
2
−7
=
√
16−7
=
√
9
= 3.
Exercises
1. Find the equation of the circle with given centre and radius:
(a) centre(3,5), radius 3; (b) centre(−2,3), radius 1; (c) centre(−1,−3), radius 2;
(d) centre(2,−2), radius 5; (e) centre(0,5)radius 4.
2. Identify the centre and radius of the following circles:
(a)x
2
+y
2
−2x−4y−20 = 0, (b)x
2
+y
2
−4x+ 6y+ 4 = 0,
(c)x
2
+y
2
+ 2x−3 = 0, (d) x
2
+y
2
+ 6x+ 7y−14
3
4
= 0,
(e)3x
2
+ 3y
2
−6x+ 9y+ 5 = 0.
4. The equation of a tangent to a circle at a given point
What is the equation of the tangent to the circlex
2
+y
2
+ 2x+ 4y−3 = 0at the point(1,−4)
on the circle?
For a question like this, we should check first that the given point does indeed lie on the circle.
If we substitutex= 1andy=−4into the equation, we obtain
1
2
+ (−4)
2
+ 2×1 + 4×(−4)−3 = 1 + 16 + 2 + (−16)−3 = 0,
and so the equation is satisfied. In fact we can see this from a diagram.
(1, −4)
(−1, −2)
tangent
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8. We have also marked the centre of the circle on the diagram. Tofind the centre, we note that
g= 1andf= 2, so that the centre is at(−1,−2).
A tangent is a straight line that just touches the circle. To find the equation of a straight line,
we need to know either two points on it, or one point on it together with its gradient. In this
example, we know one point on the line, the point(1,−4)where it is to touch the circle. But
we do not know another point. Nor do we know the gradient. So what should we do?
One fact we do know is that the tangent to a circle is perpendicular to the radius at the point
of contact. In this case, we know the point of contact(1,−4), and we also know the centre
(−1,−2). We can therefore calculate the gradient of the radius from the centre to the point of
contact, and hence the gradient of the tangent.
Now the gradientmof a straight line joining the points(x1, y1)and(x2, y2)is given by
m=
y2−y1
x2−x1
.
So if we take(x1, y1) = (1,−4)and(x2, y2) = (−1,−2), the gradientm1of the radius is
m1=
(−2)−(−4)
(−1)−1
=
2
−2
=−1.
We now use the result that, if two lines with gradientsm1andm2are perpendicular, then
m1m2=−1. Here, the gradient of the radius ism1=−1, and so the gradient of the tangent
must bem2= 1.
Now we have enough information to find the equation of the tangent. We know that the equation
of a straight line with a given gradientm= 1and containing a given point(x1, y1) = (1,−4)can
be found from the formula
y−y1=m(x−x1),
and so the equation of the tangent is given by
y−(−4) = 1×(x−1)
y+ 4 =x−1
y=x−5.
Key Point
To find the equation of the tangent to a circle through a given point of contact, you should first
find the centre of the circle and then calculate the gradientm1of the line joining the centre to
the point of contact.
Having done this, you should find the gradientm2of the tangent, using the formulam1m2=−1.
As you now know the gradient and one point on the tangent, you can find the equation of the
tangent.
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9. Example
Find the equation of the tangent at the point(0,2)to the circlex
2
+y
2
−4x+ 2y−8 = 0.
Solution
We start by finding the centre of the circle. From the equation, we see thatg=−2andf= 1,
so the centre of the circle is at(2,−1).
(0, 2)
(2, −1)
Let us take(x1, y1) = (0,2)and(x2, y2) = (2,−1). Then the gradientm1of the radius joining
these two points is
m1=
y2−y1
x2−x1
=
(−1)−2
2−0
=−
3
2
.
If the tangent has gradientm2then we must havem1m2=−1as the tangent and the radius
are perpendicular, and som2=
2
3
.
Now we can find the equation of the tangent. We know the gradientm2=
2
3
, and we know a
point(x1, y1) = (0,2). So the tangent is given by
y−y1=m2(x−x1)
y−2 =
2
3
(x−0)
y=
2
3
x+ 2.
Note that they-intercept of this line is 2, as we would expect from the fact that it passes through
the given point(0,2).
Exercises
3. Find the equation of the tangent to each circle at the pointspecified:
(a) circlex
2
+y
2
−2x−4y−20 = 0, point(4,−2);
(b) circlex
2
+y
2
+ 4x+ 2y−20 = 0, point(1,3);
(c) circlex
2
+y
2
−6x+ 4y−87 = 0, point(−3,−10);
(d) circlex
2
+y
2
+ 18x−88 = 0, point(3,5);
(e) circlex
2
+y
2
−6y−160 = 0, point(12,8).
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10. 4. Find the points of intersection of the liney= 2x+ 1and the circlex
2
+y
2
−2y+ 4 = 0.
Show that the liney= 2x+ 1is a diameter of the circle. Find the equation of the tangent to
the circle at one of the points of intersection.
5. Find the points of intersection of the liney=x−3and the circlex
2
+y
2
−2x+ 2y+ 1 = 0.
What are the tangents at the points of intersection? Where dothey intersect?
6. Find the points where the circlex
2
+y
2
−10x−10y+ 40 = 0and the liney+ 2x= 10
intersect. Find the equation of the tangent to the circle at each of the points of intersection.
Find the point of intersection of these two tangents.
7. Show that the equation of the tangent at the point(x1, y1)on the circle
x
2
+y
2
+ 2gx+ 2fy+c= 0
is given by
xx1+yy1+g(x+x1) +f(y+y1) +c= 0.
Answers
1.
(a)x
2
+y
2
−6x−10y+ 25 = 0(b)x
2
+y
2
+ 4x−6y+ 12 = 0
(c)x
2
+y
2
+ 2x+ 6y+ 6 = 0(d)x
2
+y
2
−4x+ 4y−17 = 0
(e)x
2
+y
2
−10y+ 9 = 0
2.
(a) centre(1,2), radius 5 (b) centre(2,−3), radius 3 (c) centre(−1,0), radius 2
(d) centre(−3,−
7
2
), radius 6 (e) centre(1,−
3
2
), radius
q
19
12
3.
(a)4y= 3x−20(b)4y+ 3x= 15(c)4y+ 3x+ 49 = 0
(d)5y+ 12x= 61(e)5y+ 12x= 184
4. The points of intersection are(1,3)and(−1,−1). The mid-point of these is(0,1)which is
the centre of the circle, and hencey= 2x+ 1is a diameter. The tangents are2y+x= 7and
2y+x= 3respectively.
5. The points of intersection are(1,−2)and(2,−1). The tangents arey=−2andx= 2
respectively. They intersect at the point(2,−2).
6. The points of intersection are(4,2)and(2,6). The tangents are3y+x= 10andy= 3x
respectively. They intersect at the point(1,3).
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